نمونه سوالات تستی فصل 2 ریاضی نهم همراه با جواب تشریحی

\(\frac{{16}}{5} = 3/2\)

\(\begin{array}{l}\frac{{a + \frac{1}{{a + \frac{1}{a}}}}}{a} = \frac{a}{a} + \frac{{\frac{1}{{a + \frac{1}{a}}}}}{a} = 1 + \frac{1}{{a(a + \frac{1}{a})}} = \\\\ \Rightarrow 1 + \frac{1}{{{a^2} + 1}} = 1 + \frac{1}{{{{(\frac{1}{3})}^2} + 1}} = 1 + \frac{1}{{\frac{1}{9} + 1}} = \\\\ \Rightarrow 1 + \frac{1}{{\frac{{10}}{9}}} = 1 + \frac{9}{{10}} = 1 + 0/9 = 1/9\end{array}\)

\(\begin{array}{l}x = 0/\overline 9 \Rightarrow 10x = 9/\overline 9 \Rightarrow 10x - x = \\\\9/\overline 9 - 0/\overline 9 = 9 \Rightarrow 9x = 9 \Rightarrow x = 1\end{array}\)

بررسی گزینه ها:

الف) نادرست؛ مثال نقض: \(\sqrt 2 \times \sqrt 3 = \sqrt 6 \)

ب) نادرست؛ مثال نقض: \(\sqrt 2 + \sqrt 3 \)  که عددی گویا نیست.

پ) درست

\(\begin{array}{l}x = 2/\overline 7 \Rightarrow 10x = 27/\overline 7 \Rightarrow 10x - x = \\\\ \Rightarrow 27/\overline 7 - 2/\overline 7 = 25 \Rightarrow 9x = 25\\\\ \Rightarrow x = \frac{{25}}{9} \Rightarrow {(\frac{a}{b})^2} = \frac{{25}}{9} \Rightarrow \frac{a}{b} = \sqrt {\frac{{25}}{9}} = \frac{5}{3}\\\\ \Rightarrow \left\{ \begin{array}{l}a = 5\\b = 3\end{array} \right. \Rightarrow a + 2b = 11\end{array}\)

\(\begin{array}{l}\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \cdots + \frac{{n - 1}}{n} = x\\\\ \Rightarrow \times {n^2}\\\\ \Rightarrow \frac{{{n^2}}}{2} + \frac{{2{n^2}}}{3} + \frac{{3{n^2}}}{4} + \cdots + \frac{{{n^2}(n - 1)}}{n} = {n^2}x\end{array}\)

\(\begin{array}{l}\sqrt {16} < \sqrt {24} < \sqrt {25} \Rightarrow 4 < \sqrt {24} < 5 \Rightarrow \\\\ \Rightarrow - 5 < - \sqrt {24} < - 4 \Rightarrow 0 < 5 - \sqrt {24} < 1\end{array}\)

\(A = \{ x|x \in \mathbb{R}, - 2 \le x < 2\} = [ - 2,2)\)

\(\begin{array}{l}\left. \begin{array}{l}x = 0/\overline {123} \\\\1000x = 123/\overline {123} \end{array} \right\} \Rightarrow 1000x - x = 123\\\\ \Rightarrow 999x = 123 \Rightarrow x = \frac{{123}}{{999}} = \frac{{41}}{{333}}\end{array}\)

\(\begin{array}{l}x = 0/\overline 3 \Rightarrow 10x = 3/\overline 3 \\\\ \Rightarrow 10x - x = 3/\overline 3 - 0/\overline 3 = 3 \Rightarrow 9x = 3\\\\ \Rightarrow x = \frac{3}{9} \Rightarrow x = \frac{1}{3} \Rightarrow 0/\overline 3 + {(0/\overline 3 )^3} = \\\\\frac{1}{3} + {(\frac{1}{3})^3} = \frac{1}{3} + \frac{1}{{27}} = \frac{{9 + 1}}{{27}} = \frac{{10}}{{27}} = 0/370 \cdots \\\\ = 0/\overline {370} \end{array}\)

\(\left. \begin{array}{l}a > 0 \Rightarrow |a| = a\\\\b < 0 \Rightarrow |a| = - b\end{array} \right\} \Rightarrow |a| + |b| = a - b\)

\(\begin{array}{l}|5x - 2| + |3x - 1| + 4 - 10x = \\\\ \Rightarrow | - 27| + | - 16| + 4 + 50 = \\\\ \Rightarrow 27 + 16 + 4 + 50 = 97\end{array}\)

\(\begin{array}{l}a < b \Rightarrow a - b < 0 \Rightarrow |a - b| = \\\\ \Rightarrow - (a - b) = b - a\\\\a < 0 \Rightarrow - a > 0 \Rightarrow 3 - a > 0 = \\\\ \Rightarrow |3 - a| = 3 - a\\\\b > 0 \Rightarrow b + 1 > 0 \Rightarrow |b + 1| = b + 1\\\\ \Rightarrow |a - b| + |3 - a| + |b + 1| = \\\\ \Rightarrow b - a + 3 - a + b + 1 = - 2a + 2b + 4\end{array}\)

\(\begin{array}{l}|b| > |a| \Rightarrow b + a > 0 \Rightarrow |a + b| + |a| + |b| = \\\\ \Rightarrow a + b - a + b = 2b\end{array}\)

وقتی قدر مطلق برابر منفی عبارت داخل آن شده، یعنی \({m^n}\)  منفی بوده است؛ بنابراین n عددی فرد و m عددی منفی می باشد.

وقتی فاصله دو عدد x و y روی محور اعداد برابر \(\sqrt 3 \)  می باشد، یعنی:

\(\begin{array}{l}|x - y| = |y - x| = \sqrt 3 \\\\| - 3(x - y)| = | - 3||x - y| = 3\sqrt 3 \end{array}\)

برای گزینه 4 ، مثال نقض می آوریم:

\(\begin{array}{l}\left\{ \begin{array}{l}x = 1 - \sqrt 3 \\\\y = 1 + \sqrt 3 \end{array} \right.\left\{ \begin{array}{l}\sqrt {{x^2}} - \sqrt {{y^2}} = |x| - |y| = \\\\|x - y| = |(1 - \sqrt 3 ) - (1 + \sqrt 3 )|\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l}|1 - \sqrt 3 | - |1 + \sqrt 3 | = (\sqrt 3 - 1) - (\sqrt 3 + 1) = - 2\\\\|2| = 2\end{array} \right.\\\\ \Rightarrow - 2 \ne 2\end{array}\)

\(\begin{array}{l}x < 2 \Rightarrow 2x < 4 \Rightarrow 2x - 4 < 0 \Rightarrow |2x - 4 = 4 - 2x\\\\x < 2 \Rightarrow 3x < 6 = 3x - 6 < 0 \Rightarrow |3x - 6| = 6 - 3x\\\\x < 2 \Rightarrow 5x < 10 \Rightarrow 5x < 16 \Rightarrow 16 - 5x > 0\\\\ \Rightarrow |16 - 5x| = 16 - 5x\\\\ \Rightarrow |2x - 4| + |3x - 6| + |16 - 5x| = \\\\ \Rightarrow (4 - 2x) + (6 - 3x) + (16 - 5x) = 26 - 10x\end{array}\)

برای هر \(x,y \in \mathbb{R}\)  داریم:

\(\left\{ \begin{array}{l}|x| - |y| \le |x + y| \le |x| + |y|\\\\|x| - |y| \le |x - y| \le |x| + |y|\end{array} \right.\)

\(\left\{ \begin{array}{l}x > 0,y > 0 \Rightarrow \frac{{|x|}}{x} = \frac{{|y|}}{y} = 1\\\\ \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = 1 + 1 = 2\\\\x < 0,y < 0 \Rightarrow \frac{{|x|}}{x} = \frac{{|y|}}{y} = - 1\\\\ \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = ( - 1) + ( - 1) = - 2\\\\x > 0,y < 0 \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = 1 + ( - 1) = 0\\\\x < 0,y > 0 \Rightarrow \frac{{|x|}}{x} + \frac{{|y|}}{y} = ( - 1) + 1 = 0\end{array} \right.\)

\(\begin{array}{l}\sqrt {{a^2} + 9 + \sqrt {36{a^2}} } = \sqrt {{a^2} + 9 + |6a|} \Rightarrow a > 0\\\\ \Rightarrow \sqrt {{a^2} + 9 + 6a} = |a + 3| = a + 3\end{array}\)